Prove that $\lim\limits_{n\rightarrow\infty} \left(1+\frac{1}{a_{n}} \right)^{a_{n}}=e$ if $\lim\limits_{n\rightarrow\infty} a_{n}=\infty$

Hint:

$$\left(1 + \frac{1}{\lfloor a_n \rfloor+1} \right)^{\lfloor a_n \rfloor} \leqslant \left(1 + \frac{1}{a_n} \right)^{a_n} \leqslant \left(1 + \frac{1}{\lfloor a_n \rfloor} \right)^{\lfloor a_n \rfloor+1}, $$

and

$$\left(1 + \frac{1}{n+1} \right)^n, \left( 1 + \frac{1}{n} \right)^{n+1} \to e$$

Given that $\lim_{n\to\infty }b_n = \lim_{n\to\infty }\frac{1}{a_{n}}=0$ we have,

$$\lim_{n \to\infty} \left(1 + \frac{1}{a_{n}} \right) ^ {a_{n} } = \lim_{n \to\infty} \exp\left(\frac{\ln\left(1 + \frac{1}{a_{n}} \right)}{\frac1{a_{n}}} \right)= \lim_{h \to0} \exp\left(\frac{\ln\left(1 +h \right)}{h} \right)=e$$

similarly

$$\lim_{n \to\infty} \left(1 +b_n \right) ^ { \frac{1}{b_{n}}} = \lim_{n \to\infty} \exp\left(\frac{\ln\left(1 + b_n\right)}{b_n} \right)= \lim_{h \to0} \exp\left(\frac{\ln\left(1 +h \right)}{h} \right)=e$$

For $|t|<1$, note that $t-{1 \over 2} t^2 \le \log(1+t) \le t$ and so $|\log(1+t)-t| \le {1 \over 2} t^2$.

Hence for $|x|>1$ we have $|\log(1+{1 \over x})-{1 \over x}| \le {1 \over 2} ({1\over x})^2$ and so $|x\log(1+{1 \over x})-1| \le {1 \over 2} {1\over |x|}$.

Hence $\lim_{x \to \infty} x\log(1+{1 \over x}) =\lim_{x \to \infty} \log(1+{1 \over x})^x = 1$ from which it follows that $\lim_{x \to \infty} (1+{1 \over x})^x = e$.

One does L'Hospital to show that $\lim_{x\rightarrow\infty}\left(1+\dfrac{1}{x}\right)^{x}=e$, then one does sequential characterisation of limit. Similar reasoning applied to $b_{n}\rightarrow 0$ case.

Let's do the 1st one ...

One of the logarithmic inequalities says: $$\frac{x}{1+x} < \ln (1 + x) < x, \forall x > -1$$

Because $\lim\limits_{n\rightarrow \infty} a_n \rightarrow \infty$, then $a_n > 0$ and $\frac{1}{a_n} > 0$ from some $n$ onwards. So $$0<\frac{1}{a_n+1}=\frac{\frac{1}{a_n}}{1+\frac{1}{a_n}} < \ln\left(1 + \frac{1}{a_n}\right) < \frac{1}{a_n}$$ Given $e^x$ is ascending: $$1<e^{\frac{1}{a_n+1}} < 1 + \frac{1}{a_n} < e^{\frac{1}{a_n}}$$ and from some $n$ onwards $$e^{\frac{a_n}{a_n+1}} < \left(1 + \frac{1}{a_n}\right)^{a_n} < e$$ Squeeze theorem finishes the proof.